Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $q = \dfrac{z^2 + 7z}{z^2 + 8z + 7} \times \dfrac{z + 1}{z - 9} $
Solution: First factor the quadratic. $q = \dfrac{z^2 + 7z}{(z + 1)(z + 7)} \times \dfrac{z + 1}{z - 9} $ Then factor out any other terms. $q = \dfrac{z(z + 7)}{(z + 1)(z + 7)} \times \dfrac{z + 1}{z - 9} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ z(z + 7) \times (z + 1) } { (z + 1)(z + 7) \times (z - 9) } $ $q = \dfrac{ z(z + 7)(z + 1)}{ (z + 1)(z + 7)(z - 9)} $ Notice that $(z + 7)$ and $(z + 1)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ z(z + 7)\cancel{(z + 1)}}{ \cancel{(z + 1)}(z + 7)(z - 9)} $ We are dividing by $z + 1$ , so $z + 1 \neq 0$ Therefore, $z \neq -1$ $q = \dfrac{ z\cancel{(z + 7)}\cancel{(z + 1)}}{ \cancel{(z + 1)}\cancel{(z + 7)}(z - 9)} $ We are dividing by $z + 7$ , so $z + 7 \neq 0$ Therefore, $z \neq -7$ $q = \dfrac{z}{z - 9} ; \space z \neq -1 ; \space z \neq -7 $